description/proof of that simplex interior of affine simplex is open on affine simplex with canonical topology
Topics
About: vectors space
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of simplex interior of affine simplex.
- The reader knows a definition of subspace topology of subset of topological space.
- The reader knows a definition of canonical \(C^\infty\) atlas for finite-dimensional real vectors space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that the simplex interior of any affine simplex is open on the affine simplex with the canonical topology.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(V\): \(\in \{\text{ the } d \text{ -dimensional real vectors spaces }\}\) with the canonical topology
\(\{p_0, ..., p_n\}\): \(\subseteq V\), \(\in \{\text{ the affine-independent sets of base points on } V\}\)
\([p_0, ..., p_n]\): \(= \text{ the affine simplex }\) as the topological subspace of \(V\)
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Statements:
\([p_0, ..., p_n]^\circ \in \{\text{ the open subsets of } [p_0, ..., p_n]\}\).
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2: Natural Language Description
For any \(d\)-dimensional real vectors space, \(V\), with the canonical topology, any affine-independent set of base points on \(V\), \(\{p_0, ..., p_n\} \subseteq V\), and the affine simplex, \([p_0, ..., p_n]\), as the topological subspace of \(V\), the simplex interior, \([p_0, ..., p_n]^\circ\) is an open subset of \([p_0, ..., p_n]\).
3: Proof
\([p_0, ..., p_n]^\circ = [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\), where \(\{F_j \vert j \in J\}\) is the set of the proper faces of \([p_0, ..., p_n]\).
When \(n = 0\), \([p_0]^\circ = [p_0]\), which is open on \([p_0]\).
Let us suppose that \(1 \le n\).
Let \((V, \phi)\) be the canonical chart with respect to any basis of \(V\), \(\{p_1 - p_0, ..., p_n - p_0, b_{n + 1}, ..., b_d\} \subseteq V\).
For any \(p' \in [p_0, ..., p_n]\), \(p' = \sum_{j \in \{0, ..., n\}} t'^j p_j = \sum_{j \in \{0, ..., n\}} t'^j (p_j - p_0) + \sum_{j \in \{0, ..., n\}} t'^j p_0 = \sum_{j \in \{1, ..., n\}} t'^j (p_j - p_0) + p_0\), which means that \(\phi (p') = (t'^1 + p^1_0, ..., t'^n + p^n_0, p^{n + 1}_0, ..., p^d_0)\), where \(\phi (p_0) = (p^1_0, ..., p^d_0)\).
Let \(p \in [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\) be any. \(\phi (p) = (t^1 + p^1_0, ..., t^n + p^n_0, p^{n + 1}_0, ..., p^d_0)\).
Let us take \(\delta\) such that \(0 \lt \delta \lt min (t^1, ..., t^n, (1 - \sum_{j \in \{1, ..., n\}} t^j) / n, 1 - t^1, ..., 1 - t^n, (\sum_{j \in \{1, ..., n\}} t^j) / n)\): \(0 \lt min (t^1, ..., t^n, (1 - (\sum_{j \in \{1, ..., n\}} t^j)) / n, 1 - t^1, ..., 1 - t^n, (\sum_{j \in \{1, ..., n\}} t^j) / n)\), because \(p\) is not on any proper face of \([p_0, ..., p_n]\) (the condition of being on a proper face is that \(t^j = 0\) for a \(j \in \{0, ..., n\}\), while \(t^0 = 1 - (\sum_{j \in \{1, ..., n\}} t^j)\), and \(t^j = 1\) implies that \(t^k = 0\) for \(k \neq j\)). Let us take \(U_{\phi (p)} = (t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times (p^{n + 1}_0 - \delta, p^{n + 1}_0 + \delta) \times ... \times (p^d_0 - \delta, p^d_0 + \delta) \subseteq \mathbb{R}^d\), an open neighborhood of \(\phi (p)\) on \(\mathbb{R}^d\).
\(U_{\phi (p)} \cap \phi ([p_0, ..., p_n])\), an open neighborhood of \(\phi (p)\) on \(\phi ([p_0, ..., p_n])\), is \((t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times \{p^{n + 1}_0\} \times ... \times \{p^d_0\}\), because \(\delta\) has been chosen such that \(0 \lt t^j - \delta \lt t^j + \delta \lt 1\) and \(0 \lt 1 - (\sum_{j \in \{1, ..., n\}} (t^j + \delta)) = 1 - (\sum_{j \in \{1, ..., n\}} t^j) - n \delta \lt 1 - (\sum_{j \in \{1, ..., n\}} (t^j - \delta)) = 1 - (\sum_{j \in \{1, ..., n\}} t^j) + n \delta \lt 1\): for any \(v = (t'^1 + p^1_0, ..., t'^n + p^n_0, p^{n + 1}_0, p^d_0) \in (t^1 + p^1_0 - \delta, t^1 + p^1_0 + \delta) \times ... \times (t^n + p^n_0 - \delta, t^n + p^n_0 + \delta) \times \{p^{n + 1}_0\} \times ... \times \{p^d_0\}\), \(t^j + p^j_0 - \delta \lt t'^j + p^j_0 \lt t^j + p^j_0 + \delta\), which implies that \(0 \lt t^j - \delta \lt t'^j \lt t^j + \delta \lt 1\) and \(\sum_{j \in \{1, ..., n\}} t^j - n \delta \lt \sum_{j \in \{1, ..., n\}} t'^j \lt \sum_{j \in \{1, ..., n\}} t^j + n \delta\), which implies that \(0 \lt 1 - \sum_{j \in \{1, ..., n\}} t^j - n \delta \lt 1 - \sum_{j \in \{1, ..., n\}} t'^j \lt 1 - \sum_{j \in \{1, ..., n\}} t^j + n \delta \lt 1\), and so, there is \(p' = \sum_{j \in \{1, ..., n\}} t'^j (p_j - p_0) + p_0 \in [p_0, ..., p_n]\) such that \(\phi (p') = v\). \(U_{\phi (p)} \cap \phi ([p_0, ..., p_n]) \subseteq \phi ([p_0, ..., p_n] \setminus \cup_{j \in J} F_j)\), because being on an \(F_j\) means that \(t'^k = 0\) for a \(k \in \{0, ..., n\}\), but we already know that \(0 \lt t'^j \lt 1\) for \(j \in \{0, ..., n\}\).
As \(\phi\) is a homeomorphism, \(\phi^{-1} (U_{\phi (p)} \cap \phi ([p_0, ..., p_n]))\) is an open neighborhood of \(p\) on \([p_0, ..., p_n]\) such that \(\phi^{-1} (U_{\phi (p)} \cap \phi ([p_0, ..., p_n])) \subseteq [p_0, ..., p_n] \setminus \cup_{j \in J} F_j\). So, \([p_0, ..., p_n] \setminus \cup_{j \in J} F_j\) is open on \([p_0, ..., p_n]\), by the local criterion for openness.
